Is there a closed form for the infinite sum $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!} \mathrm{?}$$ where a is an integer greater than or equal to $0$.
When $a=0$, the sum is just the series expansion for $e$, $\sum \limits_{n=0}^{\infty} \frac{1}{n!}=e$
When $a=1$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+1)!}= \sum \limits_{n=1}^{\infty} \frac{1}{n!} =\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \dots=e-1$.
When $a=2$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+2)!}= \sum \limits_{n=2}^{\infty} \frac{1}{n!} =\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots=e-2$.
When $a=3$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+3)!}= \sum \limits_{n=3}^{\infty} \frac{1}{n!} =\frac{1}{3!}+\frac{1}{4!}+\dots=e-\frac{5}{2}$.
$\vdots$
Thus, we can generalize that $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}= \sum \limits_{n=a}^{\infty} \frac{1}{n!}.$$
Can this infinite sum be written in a closed form in terms of $a$? Following the pattern the general form would have some value, dependent on $a$, subtracted from $e$.